A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone? | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-4 Motion In A Plane



Q.17:- A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?

 

Answer:-



Length of the string, l = 80 cm = 0.8 m
Number of revolutions = 14
Time taken = 25 s
Frequency, v = Number of revolutions / Time taken  =  14 / 25 Hz
Angular frequency, ω = 2πν
= 2 × (22/7) × (14/25)  =  88 / 25 rad s-1
Centripetal acceleration, ac = ω2r
= (88/25)2 × 0.8
= 9.91 ms-2
The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.